$\overline{BC} = 10$ $\overline{AC} = {?}$ $A$ $C$ $B$ $?$ $10$ $ \sin( \angle ABC ) = \dfrac{12}{13}, \cos( \angle ABC ) = \dfrac{5}{13}, \tan( \angle ABC ) = \dfrac{12}{5}$
Solution: $\overline{AC}$ is the opposite to $\angle ABC$ $\overline{BC}$ is adjacent to $\angle ABC$ SOH CAH TOA We know the adjacent side and need to solve for the opposite side so we can use the tan function (TOA) $ \tan( \angle ABC ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\overline{AC}}{\overline{BC}}= \frac{\overline{AC}}{10} $ Since we have already been given $\tan( \angle ABC )$ , we can set up a proportion to find $\overline{AC}$ $ \tan( \angle ABC ) = \dfrac{12}{5} = \frac{\overline{AC}}{10}$ Simplify. $\overline{AC} = 24$